Question

Find the roots of the following quadratic equation, if they exist using the quadratic formula of Sridharacharya: $x+\dfrac{1}{x}=3,x\ne 0$.

Answer 1

- Hint: Solve the given expression until you get a simple expression of x. Now for x is an integer’s which comes in its range. Similarly, when x is a real number, find its range.

Complete step-by-step solution -

Shridhar Acharya formula is the quadratic formula, which is used for finding the roots of a quadratic equation, $a{{x}^{2}}+bx+c=0$.
Where $a\ne 0$ and a, b, c are real numbers and they are real coefficients of the equation.
Now, $a{{x}^{2}}+bx+c=0$, has 2 roots which are,
$x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and $x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
The above Shridhar Acharya’s rule can be proved by solving the general form of quadratic equation i.e. solving, $a{{x}^{2}}+bx+c=0$.
Now we have been given the equation, $x+\dfrac{1}{x}=3$.
Now let us make this equation, to a quadratic equation like, $a{{x}^{2}}+bx+c=0$.

& x+\dfrac{1}{x}=3 \\\ & \dfrac{{{x}^{2}}+1}{x}=3\Rightarrow {{x}^{2}}+1=3x \\\ & \therefore {{x}^{2}}-3x+1=0 \\\ \end{aligned}$$ Now the above equation is similar to $$a{{x}^{2}}+bx+c=0$$. Thus comparing them both we get, a = 1, b = -3, c = 1. Now let us put the above values on the 2 roots. We know that, $$D={{b}^{2}}-4ac$$ Thus putting the values in the above expression, we get $$D={{\left( -3 \right)}^{2}}-4\times 1\times 1=9-4=5$$ The root of the equation is given by, $$x=\dfrac{-b\pm \sqrt{D}}{2a}$$. Thus putting values, $$x=\dfrac{-\left( -3 \right)\pm \sqrt{5}}{2\times 1}=\dfrac{3\pm \sqrt{5}}{2}$$ Thus we got the two roots as, $$x=\left( \dfrac{3+\sqrt{5}}{2} \right)$$and $$\left( \dfrac{3-\sqrt{5}}{2} \right)$$. Hence the roots of the quadratic equation $$\left( \dfrac{3+\sqrt{5}}{2} \right)$$and $$\left( \dfrac{3-\sqrt{5}}{2} \right)$$. Thus we got the required answer. Note: The discriminant is the expression, $${{b}^{2}}-4ac$$, which is defined for any quadratic equation, $$a{{x}^{2}}+bx+c=0$$. Based upon the sign of expression, you can determine how many real number solutions that quadratic equation has. If D is positive then 2 unique solutions are there. If D = 0, one solution and D is negative then 2 imaginary roots. Here D is positive, so 2 unique solutions $$\left( \dfrac{3+\sqrt{5}}{2} \right)$$and $$\left( \dfrac{3-\sqrt{5}}{2} \right)$$.